OFFSET
0,1
COMMENTS
L is the infinite real symmetric matrix indexed by primes with L[p,q] = 1/lcm(p,q), decomposing as L = D + v*v^T, where v_p = 1/p and D = diag((p-1)/p^2)_p.
L is compact and positive definite on l^2; lambda_inf is its spectral radius. The finite truncations L_n (first n primes) satisfy rho(L_n) -> lambda_inf monotonically from below; see EXAMPLE.
Conjecture (numerical evidence): lambda_max(W_N')/N -> lambda_inf, where W_N'[i,j] = floor(N/lcm(p_i,p_j)) for primes p_i, p_j <= N.
REFERENCES
H. J. S. Smith, On the value of a certain arithmetical determinant, Proc. London Math. Soc. s1-7 (1875), 208-212.
LINKS
Alessandro Munari, Table of n, a(n) for n = 0..499
Keith Bourque and Steve Ligh, On GCD and LCM matrices, Linear Algebra and its Applications, Volume 174, 1992, pp. 65-74.
Shaofang Hong, On the Bourque-Ligh Conjecture of Least Common Multiple Matrices, Journal of Algebra, Volume 218, Issue 1, 1999, pp. 216-228.
FORMULA
Equals the unique solution x in (1/4, oo) of Sum_{p prime} 1/(x*p^2 - p + 1) = 1.
EXAMPLE
lambda_inf = 0.67403618319369613993666000757650845578064568836207...
rho(L_5) = 0.65821735..., rho(L_1000) = 0.67402774... (monotone from below).
h(2/3) = 1.01673... > 1, h(3/4) = 0.85509... < 1, so lambda_inf in (2/3, 3/4), where h(x) = Sum_{p prime} 1/(x*p^2 - p + 1).
MATHEMATICA
Block[{$MaxExtraPrecision=300}, RealDigits[x/.FindRoot[
Sum[(1/x)^(j+1)*Sum[Binomial[j, m]*(-1)^(j-m)*PrimeZetaP[2j+2-m], {m, 0, j}],
{j, 0, 200}]==1, {x, 2/3, 3/4}, WorkingPrecision->110], 10, 75][[1]]]
PROG
(Python)
# Newton-Raphson for Sum_{p prime} 1/(x*p^2-p+1) = 1; tail via prime zeta.
from math import comb
from mpmath import mp, mpf, primezeta
from sympy import primerange
mp.dps = 75; K, P = 15, 10**5
ps = list(primerange(2, P+1))
pzt = {s: primezeta(s)-sum(mpf(p)**(-s) for p in ps) for s in range(2, 2*K+2)}
pl = [(mpf(p), mpf(p)**2) for p in ps]
def fdf(x):
h, d, xk = sum(1/(x*q-p+1) for p, q in pl), -sum(q/(x*q-p+1)**2 for p, q in pl), mpf(1)
for k in range(K):
xk *= x; c = sum((-1)**(k-j)*comb(k, j)*pzt[2*k+2-j] for j in range(k+1))
h += c/xk; d -= (k+1)*c/(xk*x)
return h-1, d
x = mpf('0.75')
for _ in range(60):
r, dr = fdf(x); x -= r/dr
if abs(r/dr) < mpf(10)**-55: break
print(x)
(PARI) solve(x=0.6, 0.7, sumeulerrat(1/(x*p^2-p+1))-1) \\ Hugo Pfoertner, Jun 04 2026
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Alessandro Munari, Jun 02 2026
STATUS
approved
